# Why do waveforms that are symmetrical above and below their horizontal centerlines contain no even-numbered harmonics?

All About Circuits site states that waveforms that are symmetrical above and below their horizontal centerlines contain no even-numbered harmonics. Can somebody explain this mathematically, or point to a resource? I do not know much about dsp, but understand what a Fourier transform is.

• 0 Hz is an even (0th) harmonic and if you have a constant bias you will have it. To close this loophole, "horizontal centerline" can be replaced by "zero level". – Olli Niemitalo Feb 11 '19 at 6:37

The complex Fourier coefficients of a $$T$$-periodic function $$f(t)$$ are given by

$$c_n=\frac{1}{T}\int_{0}^Tf(t)e^{-j2\pi nt/T}dt\tag{1}$$

with

$$f(t)=\sum_{n=-\infty}^{\infty}c_ne^{j2\pi nt/T}\tag{2}$$

The coefficients with even indices are

\begin{align}c_{2n}&=\frac{1}{T}\int_{0}^Tf(t)e^{-j4\pi nt/T}dt\\&=\frac{1}{T}\left[\int_{0}^{T/2}f(t)e^{-j4\pi nt/T}dt+\int_{T/2}^{T}f(t)e^{-j4\pi nt/T}dt\right]\\&=\frac{1}{T}\left[\int_{0}^{T/2}f(t)e^{-j4\pi nt/T}dt+\int_{0}^{T/2}f(t+T/2)e^{-j4\pi nt/T}e^{-j2\pi n}dt\right]\\&=\frac{1}{T}\int_{0}^{T/2}\left[f(t)+f(t+T/2)\right]e^{-j4\pi nt/T}dt\\&=\frac12\frac{2}{T}\int_{0}^{T/2}\left[f(t)+f(t+T/2)\right]e^{-j2\pi nt/(T/2)}dt\\&=\frac12 d_n\tag{3}\end{align}

where $$d_n$$ are the complex Fourier coefficients of the $$T/2$$-periodic function $$g(t)=f(t)+f(t+T/2)$$. These coefficients can only be zero if $$g(t)=0$$, i.e., if $$f(t)=-f(t+T/2)$$. This latter condition is exactly the symmetry condition you mention in your question. Consequently, the even Fourier coefficients are zero if (and only if) $$f(t)$$ satisfies

$$f(t)=-f(t+T/2)\tag{4}$$

Or, in other words, a $$T$$-periodic function $$f(t)$$ has only odd harmonics if it is $$T/2$$-antiperiodic. I.e., if you shift the function by half its period and flip it across the horizontal axis, it must look the same as before.

An $$M\text{-}$$periodic waveform that has half-periods of length $$M/2$$ that are of opposite signs but of equal absolute value are by definition $$M/2\text{-}$$Bloch-periodic with a coefficient of $$-1:$$ Each period of length $$M/2$$ is identical to the previous period multiplied by the coefficient.

A generalized frequency-shifted discrete Fourier transform (DFT) of length $$M/2$$ with a frequency shift of $$0.5$$ bin widths, or $$\frac{0.5\times2\pi}{M/2}$$ in angular frequency, has extended basis functions that are $$M/2\text{-}$$Bloch-periodic with a coefficient of $$-1.$$ The $$M/2\text{-}$$Bloch-periodic waveform can be represented by a weighted sum of the extended basis functions. The basis functions have angular frequencies $$\frac{0.5\times2\pi}{M/2}, \frac{1.5\times2\pi}{M/2}, \frac{2.5\times2\pi}{M/2}, \ldots.$$ This can be rewritten as $$\frac{1\times2\pi}{M}, \frac{3\times2\pi}{M}, \frac{5\times2\pi}{M}, \ldots.$$ The extended basis functions do not contain even harmonics of the $$M\text{-}$$periodic angular frequency $$\frac{2\pi}{M}.$$ This property is retained by the waveform which is their weighted sum.

A function that is $$N\text{-}$$Bloch-periodic with coefficient $$-1$$ is also called an $$N\text{-}$$antiperiodic function.